Math Challenge - August 2019

[fresh_42]

You are hard to follow.

$2x^6+3y^6=z^3$

It suffices to prove that $2x^3+3y^3=1$ has no rational roots (expression obtained by dividing by $z^3$ ).

Done in post #89, although the additional variables make it unnecessarily complicated (see post #90).

We see that $\frac{3y^3}{2} = \frac{1}{2} -x^3$

We see that $x^3= \frac{1}{2} - t^3$ and $y= \frac{2t^3}{3}$

We do not "see" it. What we see is that you invented out of the blue a certain number $t$. What you should have done is telling us, that you set $t:= \sqrt[3]{\frac{3}{2}y} \in \mathbb{R}$ such that $y=\frac{2}{3}t^3$. But now $x^3\neq \frac{1}{2}-t^3$.

To correct this, we have to set $t:=\sqrt[3]{\frac{3}{2}}y$. Now $x^3= \frac{1}{2}-t^3$.

If t is rational, then y is irrational.

How that?

If t is irrational, then x is irrational. Hence there are no solutions for x and y in rationals.

 

[member 587159]

Proof of fact. Suppose $S\in\Sigma$ has property $(P)$. Then $\Sigma _S$ is an infinite sub sigma algebra. Pick $T \in \Sigma _S \setminus \{\emptyset, S\}$ Write $S= T\cup (S\cap T^c)$. By (E) at least one of the respective sub sigma algebras must be infinite.

First of all, sorry for the late reply. I'm trying to understand this part of your answer. I guess you want to show that either $\Sigma_T$ or $\Sigma_{S \cap T^c}$ has property $(P)$, right?

Can you please explain me why the following sentence is true?

"By (E) at least one of the respective sub sigma algebras must be infinite."

 

[nuuskur]

Can you please explain me why the following sentence is true?

"By (E) at least one of the respective sub sigma algebras must be infinite."

If $\Sigma _T \cup \Sigma _{S\cap T^c}$ was a finite set, then their generated sub sigma algebra, which is $\Sigma _S$, would also be finite.

I guess you want to show that either $\Sigma_T$ or $\Sigma_{S \cap T^c}$ has property $(P)$, right?

They can both have that property, too.

 

[Not anonymous]

8. Show that there is no triple $(a,b,c) \in \mathbb{Z}^3\setminus \{(0,0,0)\}$ such that $a^2 + b^2 = 3 c^2$.

After spending a few seconds thinking about it, I realized that this problem is much simpler than what I had imagined when I first came across this question in the challenge list. An interesting problem nevertheless for people like me who have never had the opportunity to a course in number theory and the like and want some warm-up questions before moving to tougher problems.

I think there might be a simpler proof than what I came up with.

Spoiler: A solution for question 8

Since $a^2 + b^2 = 3 c^2$, it must be the case that $a^2 + b^2 \equiv 0 \mod 3$. Since $(x + y) \mod n = x \mod n + y \mod n)$, we consider all possible combinations of $a \mod 3$ and $b \mod 3$ and find their effect on $a^2 \mod 3 + b^2 \mod 3$. We only need to consider the modulo pairs (0, 0), (0, 1), (0, 2), (1, 1), (1, 2) and (2, 2) for the modulos of a and b, since the remaining cases like (2, 1) are symmetric. And we use the observation that if $x \equiv k \mod n$, then $x^2 \equiv k^2 \mod n$ (since $x = pn + k$ for some integer $p$ and $x^2 = (p^2 n^2 + k^2 + 2kpn) \equiv k^2 \mod n$).

$$
\begin{array}{|c|c|c|c|c|}
\hline a \mod 3 & b \mod 3 & a^2 \mod 3 & b^2 \mod 3 & (a^2 + b^2) \mod 3 \\
\hline 0 & 0 & 0 & 0 & 0 \\
\hline 0 & 1 & 0 & 1 & 1 \\
\hline 0 & 2 & 0 & 1 & 1 \\
\hline 1 & 1 & 1 & 1 & 2 \\
\hline 1 & 2 & 1 & 1 & 2 \\
\hline 2 & 2 & 1 & 1 & 2 \\
\hline
\end{array}
$$

So the only case where $(a^2 + b^2) \mod 3 = 0$ is when the LHS modulo pair is (0, 0), i.e. both a and b are multiples of 3, say $a = 3A, b = 3B$ for some non-zero integers $A, B$ (since question states a, b, c are all non-zero). Using this in the original expression, we get $a^2 + b^2 = 9A^2 + 9B^2 = 3 c^2 \Rightarrow A^2 + B^2 = \frac {c^2} {3}$

Since $A^2 + B^2$ and hence $\frac {c^2} {3}$ is an integer, and $c$ is an integer and 3 is prime, it must be the case that $c$ too is a multiple of 3, say $c = 3C$ for some non-zero integer $C$. So the above equivalence becomes, $A^2 + B^2 = \frac {9C^2} {3} = 3C^2$, which is of the same for as the original equation. So as long as the 2 integers on LHS are multiples of 3, we can reduce the expression to another expression of the same form but with the absolute values of the integers becoming smaller. This cannot continue infinitely, implying that we will reach a point where at least one of the 2 integers is no longer a multiple of 3, meaning that the modulo pair is no longer (0, 0). And for any such non-(0, 0) pair, we already showed that the RHS cannot be a multiple of 3, but this contradicts the expression from that required the RHS to be a multiple of 3. Therefore, the only remaining possibility, i.e. of having an LHS module pair of the form (0, 0) that also satisfies the expression form $a^2 + b^2 = 3 c^2$, too is ruled out. Hence proved.

 

[member 587159]

8. Show that there is no triple $(a,b,c) \in \mathbb{Z}^3\setminus \{(0,0,0)\}$ such that $a^2 + b^2 = 3 c^2$.

After spending a few seconds thinking about it, I realized that this problem is much simpler than what I had imagined when I first came across this question in the challenge list. An interesting problem nevertheless for people like me who have never had the opportunity to a course in number theory and the like and want some warm-up questions before moving to tougher problems.

I think there might be a simpler proof than what I came up with.

Spoiler: A solution for question 8

Since $a^2 + b^2 = 3 c^2$, it must be the case that $a^2 + b^2 \equiv 0 \mod 3$. Since $(x + y) \mod n = x \mod n + y \mod n)$, we consider all possible combinations of $a \mod 3$ and $b \mod 3$ and find their effect on $a^2 \mod 3 + b^2 \mod 3$. We only need to consider the modulo pairs (0, 0), (0, 1), (0, 2), (1, 1), (1, 2) and (2, 2) for the modulos of a and b, since the remaining cases like (2, 1) are symmetric. And we use the observation that if $x \equiv k \mod n$, then $x^2 \equiv k^2 \mod n$ (since $x = pn + k$ for some integer $p$ and $x^2 = (p^2 n^2 + k^2 + 2kpn) \equiv k^2 \mod n$).

$$
\begin{array}{|c|c|c|c|c|}
\hline a \mod 3 & b \mod 3 & a^2 \mod 3 & b^2 \mod 3 & (a^2 + b^2) \mod 3 \\
\hline 0 & 0 & 0 & 0 & 0 \\
\hline 0 & 1 & 0 & 1 & 1 \\
\hline 0 & 2 & 0 & 1 & 1 \\
\hline 1 & 1 & 1 & 1 & 2 \\
\hline 1 & 2 & 1 & 1 & 2 \\
\hline 2 & 2 & 1 & 1 & 2 \\
\hline
\end{array}
$$

So the only case where $(a^2 + b^2) \mod 3 = 0$ is when the LHS modulo pair is (0, 0), i.e. both a and b are multiples of 3, say $a = 3A, b = 3B$ for some non-zero integers $A, B$ (since question states a, b, c are all non-zero). Using this in the original expression, we get $a^2 + b^2 = 9A^2 + 9B^2 = 3 c^2 \Rightarrow A^2 + B^2 = \frac {c^2} {3}$

Since $A^2 + B^2$ and hence $\frac {c^2} {3}$ is an integer, and $c$ is an integer and 3 is prime, it must be the case that $c$ too is a multiple of 3, say $c = 3C$ for some non-zero integer $C$. So the above equivalence becomes, $A^2 + B^2 = \frac {9C^2} {3} = 3C^2$, which is of the same for as the original equation. So as long as the 2 integers on LHS are multiples of 3, we can reduce the expression to another expression of the same form but with the absolute values of the integers becoming smaller. This cannot continue infinitely, implying that we will reach a point where at least one of the 2 integers is no longer a multiple of 3, meaning that the modulo pair is no longer (0, 0). And for any such non-(0, 0) pair, we already showed that the RHS cannot be a multiple of 3, but this contradicts the expression from that required the RHS to be a multiple of 3. Therefore, the only remaining possibility, i.e. of having an LHS module pair of the form (0, 0) that also satisfies the expression form $a^2 + b^2 = 3 c^2$, too is ruled out. Hence proved.

Looks correct. A solution anaologuous to yours was already provided though.

 

[member 587159]

If $\Sigma _T \cup \Sigma _{S\cap T^c}$ was a finite set, then their generated sub sigma algebra, which is $\Sigma _S$, would also be finite.

I still don't see how (E) implies that

$$\Sigma_S = \sigma(\Sigma_T \cup \Sigma_{S\cap T^c})$$

 

[nuuskur]

I still don't see how (E) implies that

$$\Sigma_S = \sigma(\Sigma_T \cup \Sigma_{S\cap T^c})$$

Right, I wrote $(E)$ for $S=X$ and $T=S$. Perhaps, "similarly to (E)" would have been a better choice of words. The proof is the same up to labelling, regardless. Since $\Sigma _T \cup \Sigma _{S\cap T^c} \subseteq \Sigma _S$ we have
$$\sigma \left (\Sigma _T \cup \Sigma _{S\cap T^c}\right ) \subseteq \Sigma _S.$$
Conversely, for every $A\in\Sigma_S$, we have
$$A = A \cap (T\cup S\cap T^c) = [A\cap T] \cup [A\cap (S\cap T^c)]\in \sigma \left (\Sigma _T \cup \Sigma _{S\cap T^c}\right )$$

 

[Not anonymous]

8. Show that there is no triple $(a,b,c) \in \mathbb{Z}^3\setminus \{(0,0,0)\}$ such that $a^2 + b^2 = 3 c^2$.

After spending a few seconds thinking about it, I realized that this problem is much simpler than what I had imagined when I first came across this question in the challenge list. An interesting problem nevertheless for people like me who have never had the opportunity to a course in number theory and the like and want some warm-up questions before moving to tougher problems.

I think there might be a simpler proof than what I came up with.

Spoiler: A solution for question 8

Since $a^2 + b^2 = 3 c^2$, it must be the case that $a^2 + b^2 \equiv 0 \mod 3$. Since $(x + y) \mod n = x \mod n + y \mod n)$, we consider all possible combinations of $a \mod 3$ and $b \mod 3$ and find their effect on $a^2 \mod 3 + b^2 \mod 3$. We only need to consider the modulo pairs (0, 0), (0, 1), (0, 2), (1, 1), (1, 2) and (2, 2) for the modulos of a and b, since the remaining cases like (2, 1) are symmetric. And we use the observation that if $x \equiv k \mod n$, then $x^2 \equiv k^2 \mod n$ (since $x = pn + k$ for some integer $p$ and $x^2 = (p^2 n^2 + k^2 + 2kpn) \equiv k^2 \mod n$).

$$
\begin{array}{|c|c|c|c|c|}
\hline a \mod 3 & b \mod 3 & a^2 \mod 3 & b^2 \mod 3 & (a^2 + b^2) \mod 3 \\
\hline 0 & 0 & 0 & 0 & 0 \\
\hline 0 & 1 & 0 & 1 & 1 \\
\hline 0 & 2 & 0 & 1 & 1 \\
\hline 1 & 1 & 1 & 1 & 2 \\
\hline 1 & 2 & 1 & 1 & 2 \\
\hline 2 & 2 & 1 & 1 & 2 \\
\hline
\end{array}
$$

So the only case where $(a^2 + b^2) \mod 3 = 0$ is when the LHS modulo pair is (0, 0), i.e. both a and b are multiples of 3, say $a = 3A, b = 3B$ for some non-zero integers $A, B$ (since question states a, b, c are all non-zero). Using this in the original expression, we get $a^2 + b^2 = 9A^2 + 9B^2 = 3 c^2 \Rightarrow A^2 + B^2 = \frac {c^2} {3}$

Since $A^2 + B^2$ and hence $\frac {c^2} {3}$ is an integer, and $c$ is an integer and 3 is prime, it must be the case that $c$ too is a multiple of 3, say $c = 3C$ for some non-zero integer $C$. So the above equivalence becomes, $A^2 + B^2 = \frac {9C^2} {3} = 3C^2$, which is of the same for as the original equation. So as long as the 2 integers on LHS are multiples of 3, we can reduce the expression to another expression of the same form but with the absolute values of the integers becoming smaller. This cannot continue infinitely, implying that we will reach a point where at least one of the 2 integers is no longer a multiple of 3, meaning that the modulo pair is no longer (0, 0). And for any such non-(0, 0) pair, we already showed that the RHS cannot be a multiple of 3, but this contradicts the expression from that required the RHS to be a multiple of 3. Therefore, the only remaining possibility, i.e. of having an LHS module pair of the form (0, 0) that also satisfies the expression form $a^2 + b^2 = 3 c^2$, too is ruled out. Hence proved.

For the last step of my proof, I found a simpler variant, based on same fundamentals however.

Spoiler

We need to prove that $a^2 + b^2 = 3 c^2$ where $(a,b,c) \in \mathbb{Z}^3\setminus \{(0,0,0)\}$ does not have a solution for the case of $(a \mod 3, b \mod 3) \equiv (0, 0)$ (absence of solution for other modulo combinations was already proven).

Say $a = 3^{p}z, b = 3^{q}y, c = 3^{r}z$, where $p, q, r$ are the highest powers associated with 3 in the prime factorization of $a, b, c$ respectively. By this definition, $x, y, z$ will all be non-multiples of 3, implying $x \mod 3 \neq 0, y \mod 3 \neq 0, z \mod 3 \neq 0$. If we assume $(a \mod 3, b \mod 3) \equiv (0, 0)$, then $p, q$ should be positive integers. Therefore, $a^2 + b^2 = 3 c^2 \Rightarrow 3^{2p}x^2 + 3^{2q}y^2 = 3 \times 3^{2r}z^2$. If we arbitrarily assume that $p \ge q$, then LHS becomes $3^{2p} (x^2 + 3^{2q - 2p}y^2)$. The proof of $p \ge q$ case is symmetric.

Subcase 1: If $p = q$, then LHS simplifies further to $3^{2p} (x^2 + y^2)$. Since x, y are non-multiples of 3, $(x^2 + y^2) \mod 3 \neq 0$ as already show earlier, so the exponent of 3 in prime factorization of LHS is $2p$, an even natural number.

Subcase 2: If $p \neq q$, then too exponent of 3 in prime factorization of LHS still remains $2p$, because then in $(x^2 + 3^{2q - 2p}y^2)$, the 2nd component, $3^{2q - 2p}y^2$ would be a multiple of 3 but $x^2$ would not be (because x is not a multiple of 3 by definition), and hence $(x^2 + 3^{2q - 2p}y^2) \mod 3 \neq 0$. So here too, the exponent of 3 in prime factorization of LHS is an even natural number.

Whereas, on the RHS we have $3 \times 3^{2r}z^2 = 3^{2r+1} z^2$, and by definition $z$ is a non-multiple of 3. Hence the exponent of 3 in prime factorization of RHS is $(2r + 1)$, an odd number.

This means that the exponents of 3 on LHS and RHS cannot match, which contradicts the requirement of LHS = RHS. Hence proved

 

[nuuskur]

For the sake of convenience I have put all candidate solutions to this post. If possible, please delete #73 and #76.

The fundamental theorem of abelian groups says that for any finite abelian group
$$
A \cong \mathbb Z_{p_1^{k_1}} \oplus \ldots \oplus \mathbb Z_{p_m^{k_m}}
$$
I'll be using Sylow's theorems which I will refer to as theorems 1, 2 and 3 respectively. (https://en.wikipedia.org/wiki/Sylow_theorems)

Spoiler: Result

Of order $6$ there are
$$
\mathbb Z_6 \qquad S_3.
$$
Of order $8$ there are
$$
\mathbb Z_8 \qquad \mathbb Z_4 \oplus \mathbb Z _2 \qquad \mathbb Z_2 \oplus\mathbb Z_2\oplus\mathbb Z_2\qquad D_4 \qquad \mbox{Dic}_2.
$$
Of order $12$ there are
$$
\mathbb Z_{12}\qquad \mathbb Z_6 \oplus \mathbb Z_2 \qquad A_4 \qquad \mathbb Z_3 \rtimes_\varphi \mathbb Z_4 \qquad \mathbb Z_3 \rtimes _\varphi (\mathbb Z_2\oplus\mathbb Z_2).
$$

Spoiler: Problem 5 - order 6

In the abelian case there are $\mathbb Z_6\cong\mathbb Z_2\oplus\mathbb Z_3$.

Quick remark. Being a Sylow $p-$subgroup is invariant with respect to conjugating, thus if $H$ was a unique Sylow $p-$subgroup, we would have$gHg^{-1} = H, g\in G,$ i.e $H$ would be normal.

Fact. Suppose a non-abelian group $G$ has order $pq$ (primes) with $q\equiv 1 \pmod{p}$, then there are precisely $q$ Sylow$p-$subgroups in $G$.

Proof of fact. Let $n_p,n_q$ be the numbers of Sylow $p,q-$subgroups respectively. By assumption $q>p$. By theorem 3 $n_q \equiv 1 \pmod{q}$ and $n_q\mid p$, which forces $n_q = 1$, thus we have a (normal) subgroup of order $q$, which makes it cyclic, therefore abelian.

For $n_p$ we thus have two choices: $n_p = 1,q$. If $n_p = 1$, then $G\cong \mathbb Z_p \oplus \mathbb Z_q$ would be abelian, thus it must be that $n_p = q$.

Our non-abelian group $G$ therefore contains three Sylow $2-$subgroups, call them $H_1,H_2,H_3$, which by theorem 2 are all conjugate to each other. We have $\mbox{Sym}(H_1,H_2,H_3) \cong S_3$. We show that $G\cong S_3$. Define
$$\varphi : G\to \mbox{Sym}(H_1,H_2,H_3), \quad\varphi (g)(H_j) := gH_jg^{-1}$$
(also called conjugating)

Spoiler: well-defined, morphism

Firstly, if $H$ is a Sylow $p-$subgroup, then conjugating it gives another Sylow $p-$subgroup, hence it must hold that $gH_ig^{-1} = H_j$ for some $j$.

The map $\varphi (g)$ is injective, because if $gH_ig^{-1} = gH_jg^{-1}$, then
$$h_i \in H_i \Rightarrow gh_ig^{-1} \in gH_ig^{-1} = gH_jg^{-1} \Rightarrow h_i \in H_j.$$
The argument is symmetrical, thus $H_i=H_j$. It is surjective due to finiteness.

Take $g,h\in G$, then
$$\varphi (gh) (H_j) = (gh)H_j(gh)^{-1} = g(hH_jh^{-1})g^{-1} = g (\varphi (h)(H_j)) g^{-1} = \varphi (g) (\varphi (h)(H_j)) = (\varphi (g) \varphi (h))(H_j).$$

To show it's an isomorphism, it suffices to show it's injective (because $|G| = |S_3| = 6$). Suppose $\varphi (g) = \mbox{id}$ i.e $gH_jg^{-1}=H_j$, then by definition $g\in N(H_j)$ the normaliser of $H_j$. By theorem 3 all $N(H_j) = H_j$ and $g\in H_1\cap H_2\cap H_3 = \{e\}$ i.e $g=e$, which makes $\varphi$ an isomorphism.

Spoiler: Problem 5 - order 8

In the abelian case there are
$$\mathbb Z_8 \qquad \mathbb Z_4 \oplus \mathbb Z _2 \qquad \mathbb Z_2 \oplus\mathbb Z_2\oplus\mathbb Z_2$$
Suppose $G$ is non-abelian of order $8$. If there is an element of order 8, then $G$ is cyclic, thus abelian. Suppose the maximal order is $2$ and pick $g,h\in G$. If $gh=e$, then they commute. Suppose $(gh)^2 = e$, then
$$gh = geh = g(ghgh)h = (gg)hg(hh) = ehge = hg$$
and again $G$ would be abelian. Thus we must have an element $a$ of order $4$. Its generated subgroup $\langle a \rangle =: H$ is of index $2$, therefore normal. Take $b\notin H$, then we have $bab^{-1}\in H$ due to normality. Notice that
$$(bab^{-1})^4 = bab^{-1}bab^{-1}bab^{-1}bab^{-1} = baaaab^{-1} = e.$$
If also $(bab^{-1}) ^2 = e$, then $baab^{-1} = e$ would lead to $aa=e$, contradicting the order of $a$.
Next, we find the possible values of $bab^{-1}$.

1. If $bab^{-1} = a^4 = e$, then $a=e$, which is impossible.
2. If $bab^{-1} = a$, then $ba = ab$, but this will make $G$ abelian. Indeed, pick $g,h\in G$. If they're both in $H$, then they commute. Suppose $g\notin H$ and write $g = ba^m$ and $h = a^n$, then
   $$gh = ba^ma^n = ba^na^m = a^n ba^m = hg.$$
   and if both reside outside $H$, then writing $g = ba^m$ and $h = ba^n$ would similarly lead to $gh=hg$. But our group is not abelian.
3. If $bab^{-1} = a^2$, then $(bab^{-1})^2 = e$ would contradict the order of $bab^{-1}$.

Thus, the only possibility is $bab^{-1}=a^3 =a^{-1}$.

There are two cases to consider.

1. $b$ is of order $2$, then we have $G \cong \langle a,b \mid \mbox{ord}(a) = 4, \mbox{ord}(b) = 2, bab^{-1} = a^{-1} \rangle$ this is the dihedral group $D_4$.
2. $b$ is of order $4$. We show $G$ is the dicyclic group $\mbox{Dic}_2$, where
   $$\mbox{Dic}_2 \cong \langle x,y \mid x^{4} = e, x^2 = y^2, x^{-1} = y^{-1}xy \rangle.$$
   Notice that
   $$bab^{-1} = a^{-1} \Rightarrow b^{-1}ab = b^{-1}(ba^{-1}b^{-1})b = a^{-1}.$$
   So we have left to show $a^2 = b^2$. Considering the natural projection $G\to G/H\cong\mathbb Z_2$, we have $b^2 \in H$. Consider possible values of $b^2$.
   1. $b^2 = a^4 = e$ contradicts order of $b$.
   2. $b^2 = a$ contradicts order of $a$
   3. $b^2 = a^2$ is what we want.
   4. $b^2 = a^{-1}$ contradicts order of $a^{-1}$ (hence order of $a$).
      Thus $b^2 = a^2$ must hold.

Spoiler: Problem 5 - order 12

In the abelian case there are
$$\mathbb Z_{12}\qquad \mathbb Z_6 \oplus \mathbb Z_2.$$

Suppose $G$ is non-abelian of order 12. Let $H_p$ denote a Sylow $p-$subgroup and $n_p$ the number of Sylow $p-$subgroups in $G$. Then, by theorem 1, we consider subgroups $H_2$ and $H_3$. The case $n_2=n_3=1$ yields normality and $H_2 \cong \mathbb Z_4$ or $H_2 \cong \mathbb Z_2\oplus \mathbb Z_2$ and $H_3 \cong \mathbb Z_3$ and that leads to one of the previously mentioned abelian structures.

Case 2 : $n_3= 4$. Call them $H^1,H^2,H^3,H^4$. Define
$$\varphi : G \to \mbox{Sym}(H^1,H^2,H^3,H^4) \cong S_4, \quad \varphi (g) := gH^jg^{-1}.$$
Since conjugation preserves the property of being a Sylow $p-$subgroup and injectivity of each $\varphi (g)$ is trivial, the map is well-defined. It's a routine task to check it's a morphism.

Notice, if, say, $aH^ia^{-1} = H^j$, then $aN(H^{i})a^{-1} = N(H^j)$, so taking normalisers preserves the property of being conjugate.

Now, suppose $\varphi (g) = \mbox{id}$, then by definition $g\in N(H^j)$. By theorem 3 $N(H^j) = H^j$, thus $g=e$ is forced and $\varphi$ is injective. Furthermore, each $H^j$ contains two generators, therefore $G$ has $8$ elements of order 3.

Consider the alternating group $A_4$. Trivially $\varphi (G) \cap A_4 \leqslant\varphi (G)$ and $A_4$ also contains $8$ elements of order 3, thus $G\cong A_4$.

Case 3: $n_3 = 1$. Call it $H$ and pick a Sylow $2-$subgroup $K$ (of order $4$). As $H$ is normal it holds $HK$ is a subgroup of $G$. As $H\cap K = \{e\}$ (by orders of elements), then
$$|HK| = \frac{|H||K|}{|H\cap K|} \Rightarrow HK = G.$$
We show $G\cong H\rtimes _\varphi K$, the semidirect product under $\varphi : K\to \mbox{Aut}(H)$, where
$$(h,k)\cdot (h',k') := (h\varphi (k)(h'),kk')$$
This will yield two more groups, namely the following:
$$\mathbb Z_3 \rtimes_\varphi \mathbb Z_4 \qquad \mathbb Z_3 \rtimes _\varphi (\mathbb Z_2\oplus\mathbb Z_2).$$
Define
$$\varphi : K\to \mbox{Aut}(H),\quad \varphi (k)(h) := khk^{-1}.$$
I will omit the routine checks. We show
$$\tau : G\to H\rtimes _\varphi K, \quad g=hk \mapsto (h,k),$$
is an isomorphism. Although, let's have some fun and do it the category theory way, for a change. First, we check $\tau$ is a morphism. Take $g=hk, g'=h'k'\in G$, then due to normality $kh'k^{-1} \in H$ and we get
$$\begin{align*} \tau (hkh'k') &= \tau (h(kh'k^{-1})kk') \\ &= (h(kh'k^{-1}),kk')\\ &= (h\varphi (k)(h'),kk')\\ &= (h,k)(h',k')\\ &= \tau (hk) \tau (h'k'). \end{align*}$$
Also $\tau (e) = \tau (ee) = (e,e)$. Now define
$$\omega : H\rtimes _\varphi K \to G, \quad (h,k) \mapsto hk.$$
Verify it, too, is a morphism. Take $(h,k),(h',k')\in H\rtimes _\varphi K$, then
$$\begin{align*} \omega ((h,k)(h',k')) &= \omega ((h\varphi (k)(h'), kk'))\\ &= h(kh'k^{-1})kk' \\ &= hkh'k' \\ &= \omega ((h,k)) \omega ((h',k')). \end{align*}$$
Also $\omega ((e,e)) = ee = e$. The equalities $\tau\omega = \mbox{id}_{H\rtimes _\varphi K}$ and $\omega\tau = \mbox{id}_G$ are straightforward to verify.

 

[member 587159]

Right, I wrote $(E)$ for $S=X$ and $T=S$. Perhaps, "similarly to (E)" would have been a better choice of words. The proof is the same up to labelling, regardless. Since $\Sigma _T \cup \Sigma _{S\cap T^c} \subseteq \Sigma _S$ we have
$$\sigma \left (\Sigma _T \cup \Sigma _{S\cap T^c}\right ) \subseteq \Sigma _S.$$
Conversely, for every $A\in\Sigma_S$, we have
$$A = A \cap (T\cup S\cap T^c) = [A\cap T] \cup [A\cap (S\cap T^c)]\in \sigma \left (\Sigma _T \cup \Sigma _{S\cap T^c}\right )$$

I went through your solution again and I think it is correct! Well done!

 

[nuuskur]

Spoiler: Problem 3 - work in progress

Suppose $\alpha$ is $n$-th root of unity and $\alpha \in \mathbb F_{q^m}$ for some $m\in\mathbb N$. Then by Lagrange $n \mid q^m-1$. It would suffice to compute the order of $q$ modulo $n$. As $(q,n)=1$ it holds that $n\mid q^{\phi (n)} -1$, where $\phi$ is the Euler totient map. So the order of $q$ must be a divisor of $\phi (n)$.

Not sure how to explicitly compute $m$.

 

[member 587159]

Spoiler: Problem 3 - work in progress

Suppose $\alpha$ is $n$-th root of unity and $\alpha \in \mathbb F_{q^m}$ for some $m\in\mathbb N$. Then by Lagrange $n \mid q^m-1$. It would suffice to compute the order of $q$ modulo $n$. As $(q,n)=1$ it holds that $n\mid q^{\phi (n)} -1$, where $\phi$ is the Euler totient map. So the order of $q$ must be a divisor of $\phi (n)$.

Not sure how to explicitly compute $m$.

Try to take $m$ minimal somehow.

 

[nuuskur]

Try to take $m$ minimal somehow.

Uhh, this is like trial and error. It's about reducing the powers of $x:=\phi (n) = \prod _{i\in I}p_i^{k_i}$. I can think of recursion.

I'll write the most barebones version I can think of, can likely be (heavily) optimised. Let $P(x) \Leftrightarrow n\mid q^{x}-1$.

1. Fix $a\in I$.
2. If $\neg P(x/p_a)$, then return $x$ and terminate.
3. Else put $x:= x/p_a$ and go to 1 (with $I$-many branches)

Do this for every $a\in I$ as a starting point (there will be a lot of branching). Eventually, pick the smallest $x$ that was returned, this will be the order of $q$ modulo $n$.

 

[member 587159]

Try to take $m$ minimal somehow.

I'm not looking for a closed, explicit form of the order of q mod n. Just explain why it is finite.

 

[nuuskur]

I'm not looking for a closed, explicit form of the order of q mod n. Just explain why it is finite.

Well, per assumption $n\mid q^{\phi (n)}-1$. Thus the set $\{x : n\mid q^x -1\}\subset \mathbb N$ is non-empty, therefore there is smallest one.

 

[member 587159]

Well, per assumption $n\mid q^{\phi (n)}-1$. Thus the set $\{x : n\mid q^x -1\}\subset \mathbb N$ is non-empty, therefore there is smallest one.

Where did you (implicitely) use that $(q,n)=1$? And that does not solve the problem entirely of course.

 

[nuuskur]

Where did you (implicitely) use that $(q,n)=1$? And that does not solve the problem entirely of course.

$(q,n)=1$ guarantees $n\mid q^{\phi (n)}-1$ (Euler's theorem). It's not clear to me what solution is expected. I gave a way of finding the order $m$ and justified its existence. The smallest field extension with desired property would have to be $\mathbb F_{q^m}$.

Ok, maybe there's the issue of necessarily containing a subgroup of order $n$. But the multiplicative group of a finite field is cyclic, therefore there is a subgroup of any order that divides the order of the group.

 

[member 587159]

$(q,n)=1$ guarantees $n\mid q^{\phi (n)}-1$ (Euler's theorem). It's not clear to me what solution is expected. I gave a way of finding the order $m$ and justified its existence. The smallest field extension with desired property would have to be $\mathbb F_{q^m}$.

Ok, maybe there's the issue of necessarily containing a subgroup of order $n$. But the multiplicative group of a finite field is cyclic, therefore there is a subgroup of any order that divides the order of the group.

Yes, your first solution assumed there was an n-th root and then deduced a necessary condition. Your last edit shows that this necessary condition is also sufficient to have an n-th root.

For me, you solved the question succesfully!

 

[nuuskur]

Oh, no more problems remaining Well, that's it for me, folks. Likely not taking part in next month's competition. *drops the mic*

 

[member 587159]

Oh, no more problems remaining Well, that's it for me, folks. Likely not taking part in next month's competition. *drops the mic*

Other challenge threads still have open problems.

 

[QuantumQuest]

2. Find the equation of a curve such that $y′′$ is always $2$ and the slope of the tangent line is $10$ at the point $(2,6)$.

Spoiler: Question 2

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The way you solve it is correct but there are some mistakes - most likely typos, in the solution. In any case it is already solved by @KnotTheorist.

 

[QuantumQuest]

. The maximum value of $f$ with $f(x) = x^a e^{2a - x}$ is minimal for which values of positive numbers $a$ ?

Spoiler: Question 1

undefined

Your way of thinking is correct. The question has already been solved by @Pi-is-3.

 

[Kyle Nemeth]

My attempt at 14 (I'm a novice so bare with me!)

Spoiler: Question 14

Consider a solution $(a,b,c)$ to the equation $$2x^6+3y^6=z^3$$ where $a,b,c \neq 0$ and $a,b,c \in \mathbb{Q}$. This yields $$2a^6+3b^6=c^3$$ $$\implies 2\frac {a^6}{c^3}+3 \frac{b^6}{c^3}=1$$ $$ \implies 2 \left( \frac {a^2}{c} \right)^3 +3 \left( \frac {b^2}{c} \right)^3=1$$ We search for the possibility of a rational solution. If $$ \left( \frac {a^2}{c} \right)^3 \neq \frac {r^3}{t^3}$$ for $r,t \in \mathbb {Z}$, then $\sqrt[3] \frac {r}{t} \in \mathbb {I}$ $\forall r,t$ with no common factor, due to the prime factorization theorem. This would imply that either $a \in \mathbb {I}$ or $c \in \mathbb {I}$. The same logic follows for $\left( \frac {b^2}{c} \right)^3$. So for a rational solution to exist, it must be that $$\frac {a^2}{c} = \frac {r}{t}, \frac {b^2}{c} = \frac {s}{t}$$ where $s \in \mathbb {Z}$, and we have reduced $\frac{r}{t}$ and $\frac{s}{t}$ to simplest form, as can be done with any rational number. Then $$2 \left( \frac{r^3}{t^3} \right)+3 \left( \frac {s^3}{t^3} \right) = 1$$ $$\implies 2r^3 + 3s^3=t^3$$ With algebra, we can show that $$f(s,t)=\sqrt[3] \frac{t^3-3s^3}{2}$$ If $t,s$ are both even, then $\frac {s}{t}$ is not fully simplified, but we assumed that we did simplify $\frac {s}{t}$ previously, so this cannot be the case. If $t,s$ are even and odd respectively, or vice versa, then the numerator $t^3-3s^3$ is odd. By definition, an odd number does not contain a factor of 2 and so the numerator and denominator of $f(s,t)$ will be uniquely irrational (irrational with no common factor) and so $r$ will be irrational in this case, implying $a\in \mathbb{I}$ or $c\in \mathbb{I}$. If $s,t$ are both odd, then $t^3-3s^3$ is even and the numerator may contain a factor of 2. By the prime factorization theorem, we may write $$t^3-3s^3=2^k(P_1^{n_1}\cdot P_2^{n_2}\cdot P_3^{n_3}\cdot {\dots} \cdot P_i^{n_i} \cdot {\dots})$$ where $k,n_i \in \mathbb {N}$ and $P_i$ is a prime number. Here, $2^k$ can be a perfect cube, but how do we know that $(P_1^{n_1}\cdot P_2^{n_2}\cdot P_3^{n_3}\cdot {\dots} \cdot P_i^{n_i} \cdot {\dots})$ is never a perfect cube (which may imply a possible rational solution)? Consider the case where $s=t$. Then $$r = f(s,t)=\sqrt[3] \frac {t^3-3t^3}{2}=\sqrt[3] \frac {-2t^3}{2}=-\sqrt[3] {t^3} $$ $$\implies r^3=-t^3 $$ From this we see that in the case where $s=t$, $(P_1^{n_1}\cdot P_2^{n_2}\cdot P_3^{n_3}\cdot {\dots} \cdot P_i^{n_i} \cdot {\dots})$ must be a perfect cube, since $2r^3=t^3-3s^3$ (and it must be odd as well, since any factor of 2 from the prime factorization is contained within $2^k$). However, if $r=-t$ then we get a complex set of solutions from $\frac {a^2}{c}$ and $\frac {b^2}{c}$. It can be shown for the equation $2r^3+3s^3=t^3$ that there is a 1-1 correspondence over the rationals for three separate cases: $$1) r = f(s,t)$$ $$2) s=g(r,t)$$ $$3) t=h(s,r)$$ Thus, when $s \neq t$, it cannot be true that $(P_1^{n_1}\cdot P_2^{n_2}\cdot P_3^{n_3}\cdot {\dots} \cdot P_i^{n_i} \cdot {\dots})$ is a perfect cube, which implies that in this case $r\in \mathbb{I}$ regardless of whether $2^k$ is a perfect cube or not. This would then imply that $a \in \mathbb{I}$ or $c \in \mathbb{I}$. Therefore, for any $s,t$, there cannot exist a possible rational solution. Similar cases can be proven with $g(r,t)$ and $h(s,r)$ based on the 1-1 correspondence. There only exists a trivial rational solution, $x=y=z=0$ since $$2(0)^6+3(0)^6=(0)^3$$ $$\implies 0=0$$

 

[fresh_42]

My attempt at 14 (I'm a novice so bare with me!)

Sure. You put much effort in it, I appreciate that, and it is not an easy problem. Nevertheless I'll have to be strict.

Consider a solution $(a,b,c)$ to the equation $$2x^6+3y^6=z^3$$ where $a,b,c \neq 0$ and $a,b,c \in \mathbb{Q}$.

And what in the other cases? We may assume that not all three of them, or even two equal zero, since then we are done. But what is in the cases $c=0$ or $a=0$?

This yields $$2a^6+3b^6=c^3$$ $$\implies 2\frac {a^6}{c^3}+3 \frac{b^6}{c^3}=1$$ $$ \implies 2 \left( \frac {a^2}{c} \right)^3 +3 \left( \frac {b^2}{c} \right)^3=1$$ We search for the possibility of a rational solution.

Now I get lost. What do you need $r,s,t$ for?

If $$ \left( \frac {a^2}{c} \right)^3 \neq \frac {r^3}{t^3}$$ for $r,t \in \mathbb {Z}$, ...

This is never true. We can always write a quotient of two quotients as one quotient.

... then $\sqrt[3] \frac {r}{t} \in \mathbb {I}$ $\forall r,t$ with no common factor, due to the prime factorization theorem. This would imply that either $a \in \mathbb {I}$ or $c \in \mathbb {I}$. The same logic follows for $\left( \frac {b^2}{c} \right)^3$. So for a rational solution to exist, it must be that $$\frac {a^2}{c} = \frac {r}{t}, \frac {b^2}{c} = \frac {s}{t}$$ where $s \in \mathbb {Z}$, and we have reduced $\frac{r}{t}$ and $\frac{s}{t}$ to simplest form, as can be done with any rational number. Then $$2 \left( \frac{r^3}{t^3} \right)+3 \left( \frac {s^3}{t^3} \right) = 1$$ $$\implies 2r^3 + 3s^3=t^3$$

Now we have a solution $r=-1, s=t=1$ but no solution for the original question!

With algebra, we can show that $$f(s,t)=\sqrt[3] \frac{t^3-3s^3}{2}$$ If $t,s$ are both even, then $\frac {s}{t}$ is not fully simplified, but we assumed that we did simplify $\frac {s}{t}$ previously, so this cannot be the case. If $t,s$ are even and odd respectively, or vice versa, then the numerator $t^3-3s^3$ is odd. By definition, an odd number does not contain a factor of 2 and so the numerator and denominator of $f(s,t)$ will be uniquely irrational (irrational with no common factor) and so $r$ will be irrational in this case, implying $a\in \mathbb{I}$ or $c\in \mathbb{I}$. If $s,t$ are both odd, then $t^3-3s^3$ is even and the numerator may contain a factor of 2. By the prime factorization theorem, we may write $$t^3-3s^3=2^k(P_1^{n_1}\cdot P_2^{n_2}\cdot P_3^{n_3}\cdot {\dots} \cdot P_i^{n_i} \cdot {\dots})$$ where $k,n_i \in \mathbb {N}$ and $P_i$ is a prime number. Here, $2^k$ can be a perfect cube, but how do we know that $(P_1^{n_1}\cdot P_2^{n_2}\cdot P_3^{n_3}\cdot {\dots} \cdot P_i^{n_i} \cdot {\dots})$ is never a perfect cube (which may imply a possible rational solution)? Consider the case where $s=t$. Then $$r = f(s,t)=\sqrt[3] \frac {t^3-3t^3}{2}=\sqrt[3] \frac {-2t^3}{2}=-\sqrt[3] {t^3} $$ $$\implies r^3=-t^3 $$ From this we see that in the case where $s=t$, $(P_1^{n_1}\cdot P_2^{n_2}\cdot P_3^{n_3}\cdot {\dots} \cdot P_i^{n_i} \cdot {\dots})$ must be a perfect cube, since $2r^3=t^3-3s^3$ (and it must be odd as well, since any factor of 2 from the prime factorization is contained within $2^k$). However, if $r=-t$ then we get a complex set of solutions from $\frac {a^2}{c}$ and $\frac {b^2}{c}$. It can be shown ...

Never, ever use this platitude! You bet it is in nine of ten cases wrong and thus needlessly embarrassing.

You were on the right track. Divisibility is the key. You should consider the remainders of a division by $7$. There are not many remainders possible for cubes.

 

[Kyle Nemeth]

And what in the other cases? We may assume that not all three of them, or even two equal zero, since then we are done. But what is in the cases $c=0$ or $a=0$?

Understood. I am quite unsure exactly of the nature of a "rational solution," whether each $a,b,c$ have to all be rational or if a rational solution could be constituted by something like $(a,b,c)$ where only $a,b \in \mathbb{Q}$ and maybe $c\in \mathbb{R}$ for example. Either way, I understand that 0 is indeed rational, and so I have left out those cases.

Now I get lost. What do you need $r,s,t$ for?

My goal was to reduce the problem down to integers instead of working with rational numbers, since then I could apply the parity arguments.

This is never true. We can always write a quotient of two quotients as one quotient.

I believe that was actually the intention of my argument. If $a,c$ are rational numbers (each an integer divided by an integer), then $\frac{a^2}{c}$ can be written as a single fraction in reduced form, $\frac{r}{t}$ where $r,t$ are integers.

Now we have a solution $r=-1, s=t=1$ but no solution for the original question!

This is a solution, but it would be a complex solution to the original equation and so I was under the presumption that a solution of this form would not be accepted as a rational solution to the original question.

Never, ever use this platitude! You bet it is in nine of ten cases wrong and thus needlessly embarrassing.

Would you mind pointing out exactly where my logic fails within this argument? I value this problem as a learning experience as well as a challenge. Thank you for the haste reply!

 

[fresh_42]

Most of what you wrote can be directly achieved. If we have an equation with quotients, we can always make it an integer equation, just multiply the quotients to get the least common multiple in the denominator.
Using the definition of a prime, namely $p$ is prime, if it is not $\pm 1$ and $p\,|\,ab$ requires $p\,|\,a$ or $p\,|\,b$ makes the cancellations you need. But the reduction to $2r^3+3s^3=t^3$ isn't the solution, since $(r,s,t)=(-1,1,1)$ is a counterexample.

I read your proof very carefully and achieved your conclusions with slightly different arguments. So maybe the proof can be corrected by the additional assumption, that all $r,s,t$ are positive. I got at least as far as $2r^3+3s^3=t^3$ with coprime pairs $(r,t)$ and $(s,t)$. But then the solution with $r=-1$ came to mind.

You could really try to pass all equations to their remainders modulo $7$ and look for common divisors.
If $2a^6+3b^6=c^3$ then the same equation holds if we substitute $a,b,c$ by their remainders of any division, esp. $7$.

An example of what I mean: $13^2=12^2+5^2$. Division by $5$ yields as remainders $9=3^2\equiv 2^2+0^2 = 4 \mod 5$, since $9$ and $4$ have the same remainder modulo a division by $5$. This works for any number. The above problem can be tackled by remainders modulo $7$. The remainders modulo seven are $\{\,0,1,2,3,4,5,6\,\}$ which is cubed $\{\,0,1,8,27,64,125,216\,\}$ with remainders $\{\,0,1,1,6,1,6,6\,\}=\{\,0,1,6\,\}$. This is a very limited selection of possibilities.

 

[Kyle Nemeth]

Understood, the substitutions I made were clearly not necessary and the fact that any equation with quotients can be written as an integer equation by multiplying the least common denominator is an approach I did not think of.

But the reduction to $2r^3+3s^3=t^3$ isn't the solution, since $(r,s,t)=(−1,1,1)$ is a counterexample.

I do not assert that this equation has no solutions, but rather any potential solution will ultimately give a solution to the original equation $$2x^6+3y^6=z^3$$ that is not rational.

So maybe the proof can be corrected by the additional assumption, that all $r,s,t$ are positive.

In order for $2x^6+3y^6=z^3$ to have a solution that is rational, isn't there a certain restriction on which values may be negative in a solution $(r,s,t)$? Since $$\frac {a^2}{c} = \frac {r}{t}, \frac {b^2}{c} = \frac {s}{t}$$ In order that $a$ or $b$ not be complex, for $c>0$, mustn't it be the case that a potential solution must exist of the forms $(r,s,t)$ where $r,s,t<0$ or $r,s,t>0$? or if $c<0$, then we must have $r,s<0$,$t>0$ or $r,s>0$, $t<0$? But if $r,s<0$ and $t>0$, we can let $r=-\alpha$, $s=-\beta$ and $t=\gamma$, for integers $\alpha,\beta,\gamma>0$, then (using $f(s,t)$ in post 113) $$ -\alpha = \sqrt[3] \frac{\gamma^3-3(-\beta)^3}{2}$$ $$\implies -\alpha=\sqrt[3] \frac{\gamma^3+3\beta^3}{2} $$ and this equation is never true, since $\alpha,\beta,\gamma>0$, and this is similarly the case for where $r,s>0$, $t<0$.

And what in the other cases? We may assume that not all three of them, or even two equal zero, since then we are done. But what is in the cases $c=0$ or $a=0$?

If $a=0$, then $$3b^6=c^3$$ $$\implies \frac {c}{b^2}=\sqrt[3] 3$$ which implies that $c$ or $b$ must be irrational, since a number $\frac {p}{q}\in \mathbb{I}$ if either $p$ or $q$ is irrational. If $b=0$, then $$2a^6=c^3$$ $$\implies \frac{c}{a^2}=\sqrt[3] 2$$ and so $c$ or $a$ must be irrational. If $c=0$, then $$2a^6+3b^6=0$$ $$\implies \frac {a}{b}=\sqrt[6] \frac {-3}{2}$$ which also implies $a$ or $b$ must be irrational (and complex).

You could really try to pass all equations to their remainders modulo $7$ and look for common divisors.

Ahh I see, I currently do not have much knowledge of modular arithmetic so that is another approach I was not aware of and I can see how this approach is far more simple than mine.

 

[fresh_42]

I will check later. I'm too tired now to say something reasonable, or to read something correctly.

We can assume $a.b,c > 0$ from the start, after the cases $c=0$ (easy) and $a=0$ (prime definition with $p=2$ if $b=0$ or symmetrically $p=3$ if $a=0$) are done. A solution with a negative number immediately also gets a solution with a positive number, and $c<0$ is impossible.
This leads to $2r^3+3s^3=t^3$ with positive integers $r,s,t$. That they may be assumed pairwise coprime is again a consequence of primality.

E.g. let us assume a prime $p$ divides $r$ and $t$. Then
\begin{align*}
p \,|\, 3s^3=t^3-2r^3 & \Longrightarrow p\,|\,3 \text{ or } p\,|\,s^3\\
\text{ (a) }p\,|\,s^3 &\Longrightarrow p\,|\,s\\
&\Longrightarrow p^3\,|\,s^3\\
&\Longrightarrow p^3 \text{ divides every term and can be cancelled }\\
\text{ (b) }p\,|\,3&\Longrightarrow p=3\\
&\Longrightarrow s^3= 9\cdot t'^3 - 2\cdot 9\cdot r'^3\\
&\Longrightarrow 3\,|\,s^3\\
&\Longrightarrow 3\,|\,s\\
&\Longrightarrow 3^3\,|\,s^3\\
&\Longrightarrow 3^3 \text{ divides every term and can be cancelled }
\end{align*}
This is the way primality is normally used. Here it allows us to assume that $r,s,t$ are pairwise coprime, since otherwise we would get a common factor cubed, which we can cancel and still have the same equation structure $2r'^3+3s'^3=t'^3$. So we continue to cancel all common factors. Now we have

• $2r^3+3s^3=t^3$
• $r,s,t > 0$
• $r,s,t$ are pairwise coprime

If $c=0$, then $2a^6+3b^6=0 \quad \implies \frac {a}{b}=\sqrt[6] \frac {-3}{2}$ which also implies $a$ or $b$ must be irrational (and complex).

You need $b\neq 0$ here, but this is not necessary. Both terms of the right hand side are always non negative. If their sum equals zero, then both summands have to be zero, i.e. $a=b=c=0$.

 

[fresh_42]

Consider a solution $(a,b,c)$ to the equation
$$
2x^6+3y^6=z^3
$$
where $a,b,c \neq 0$ and $a,b,c \in \mathbb{Q}$. This yields
$$
2a^6+3b^6=c^3
$$
$$
\implies 2\frac {a^6}{c^3}+3 \frac{b^6}{c^3}=1
$$
$$
\implies 2 \left( \frac {a^2}{c} \right)^3 +3 \left( \frac {b^2}{c} \right)^3=1
$$

I skip the part here that is a bit confusing and unnecessary. We can always transform $2a^6+3b^6=c^3$ into ...

$$
\implies 2r^3 + 3s^3=t^3
$$

where we may assume w.l.o.g. $r,s,t > 0$ and $r,s,t$ are pairwise coprime integers (cp. post #118).

With algebra, we can show that
$$
f(s,t)=\sqrt[3] \frac{t^3-3s^3}{2}
$$
If $t,s$ are both even...

Obsolete, if we first guarantee that $s$ and $t$ are coprime.

... then $\frac {s}{t}$ is not fully simplified, but we assumed that we did simplify $\frac {s}{t}$ previously, so this cannot be the case. If $t,s$ are even and odd respectively, or vice versa, then the numerator $t^3-3s^3$ is odd. By definition, an odd number does not contain a factor of 2 and so the numerator and denominator of $f(s,t)$ will be uniquely irrational (irrational with no common factor) and so $r$ will be irrational in this case, implying $a\in \mathbb{I}$ or $c\in \mathbb{I}$.

What we really have in this case is, that a number $z+\frac{1}{2}$ is under the root. We need to show that there is no integer $r$ with $r^3=z+\frac{1}{2}$. But this is true and no argument involving irrationality is needed.
Keep it simple and only use what you have, step by step.

If $s,t$ are both odd, then $t^3-3s^3$ is even and the numerator may contain a factor of 2. By the prime factorization theorem, we may write
$$
t^3-3s^3=2^k(P_1^{n_1}\cdot P_2^{n_2}\cdot P_3^{n_3}\cdot {\dots} \cdot P_i^{n_i} \cdot {\dots})
$$
where $k,n_i \in \mathbb {N}$ and $P_i$ is a prime number. Here, $2^k$ can be a perfect cube, but how do we know that $(P_1^{n_1}\cdot P_2^{n_2}\cdot P_3^{n_3}\cdot {\dots} \cdot P_i^{n_i} \cdot {\dots})$ is never a perfect cube (which may imply a possible rational solution)? Consider the case where $s=t$. Then
$$
r = f(s,t)=\sqrt[3] \frac {t^3-3t^3}{2}=\sqrt[3] \frac {-2t^3}{2}=-\sqrt[3] {t^3}
$$
$$
\implies r^3=-t^3
$$
From this we see that in the case where $s=t$...

... which again is obsolete, since $s$ and $t$ can be assumed coprime.

...$(P_1^{n_1}\cdot P_2^{n_2}\cdot P_3^{n_3}\cdot {\dots} \cdot P_i^{n_i} \cdot {\dots})$ must be a perfect cube, since $2r^3=t^3-3s^3$ (and it must be odd as well, since any factor of 2 from the prime factorization is contained within $2^k$). However, if $r=-t$ then we get a complex set of solutions from $\frac {a^2}{c}$ and $\frac {b^2}{c}$. It can be shown for the equation $2r^3+3s^3=t^3$ that there is a 1-1 correspondence over the rationals for three separate cases: $$
1) r = f(s,t)
$$
$$
2) s=g(r,t)
$$
$$
3) t=h(s,r)
$$

Summary: We are left with $s\neq t$, both odd, and
$$
\dfrac{t^3-3s^3}{2} = \left( 2^k P_1^{n_1}\cdot P_2^{n_2}\cdot \ldots \cdot P_{n_k}^{n_k}\right)^{1/3} \in \mathbb{Z} \;\;\text{ i.e. }\;\; 3\,|\, k,n_j
$$

Thus, when $s \neq t$, it cannot be true that $(P_1^{n_1}\cdot P_2^{n_2}\cdot P_3^{n_3}\cdot {\dots} \cdot P_i^{n_i} \cdot {\dots})$ is a perfect cube...

Sorry, but I do not see this. Why is it impossible that all powers are divisible by three? What happens if they were? E.g. all zero is a possibility.

...which implies that in this case $r\in \mathbb{I}$ regardless of whether $2^k$ is a perfect cube or not. This would then imply that $a \in \mathbb{I}$ or $c \in \mathbb{I}$. Therefore, for any $s,t$, there cannot exist a possible rational solution. Similar cases can be proven with $g(r,t)$ and $h(s,r)$ based on the 1-1 correspondence. There only exists a trivial rational solution, $x=y=z=0$ since
$$
2(0)^6+3(0)^6=(0)^3
$$
$$
\implies 0=0
$$

 

[Pi-is-3]

You are hard to follow.

Done in post #89, although the additional variables make it unnecessarily complicated (see post #90).

We do not "see" it. What we see is that you invented out of the blue a certain number $t$. What you should have done is telling us, that you set $t:= \sqrt[3]{\frac{3}{2}y} \in \mathbb{R}$ such that $y=\frac{2}{3}t^3$. But now $x^3\neq \frac{1}{2}-t^3$.

To correct this, we have to set $t:=\sqrt[3]{\frac{3}{2}}y$. Now $x^3= \frac{1}{2}-t^3$.

How that?

Sorry for late reply. I was not able to come online a lot since a long time. Anyway, I tried to prove it, but I couldn't. Hence my proof is wrong. Sorry.